Solving Single Variable Equations
By: Mikerline Paul
There will be five different topics: single variable equations with a variable on one side, a variable onboth sides, fractions, quadratics, and cubics.
One Step Single Variable Equations
There are different types of single variable equations so there are different ways of solving them.
For instance you can have anequation with x on one side or x on both sides. You can have a cubic or quadratic in the equation also.
Ex: 5 + x = 8
You want to get the variable (in this case x) on one side of the equal sign and a number on the other side.
So the first thing you do is subtract 5 from each side.
Now you have
5 + x =8
-5 -5
The 5 then becomes 0 and you’re just left with 8 - 5 which is 3. So the solution to the equation is x=3.
Single Variable Equations with X on Both Sides
If you have an equation with a variable on both sides there’s more to it.
The first step would be to combine like terms.
Numbers on both sides of the equations that are being multiplied by the variable are like terms and numbers that are not being multiplied by the variable on both sides of the equation are like terms. For example 4x and 7x would be like terms, but 2 and 7x would not be like terms.
Ex: 2x + 4= 6x – 2
To combine like terms inthis problem you would need to add 2 to both sides
2x + 4= 6x – 2
+2 +2
You are left with 2x + 6= 6x
Now you have to subtract 2x from both sides
2x + 6= 6x
-2x -2x
6= 4x
Now that you have isolated the variable you’re solving for, all you have to do is divide both sides by 4x to solve for x.
6/4x +4x/4x
1.5=x
After you get your answer, plug in the x value you got to make sure that you solved the problem correctly.
2 (1.5) + 4= 6(1.5) – 2
3 + 4 = 9- 2 7=7 !
*There is also another way to solve for x with an equation like this. Just graph both equations and the point at which they meet (point of intersection) is your solution.*
Single Variable Equations with Fractions
Equations don’t only have whole number in them, they have fractions as well. You solve it the same way. You just have to remember the rules of multiplying and dividing fractions. Also, if you are multiplying or dividing a fraction by a whole number don’t forget to turn it into a fraction by putting it over 1. For example 2 would be 2/1.
Ex: 1/2x+2= 1/4x -6
The easiest thing to do would be to combine the whole numbers first.
You would do this by adding 6 to both sides
1/2x+2= 1/4x -6
+6 +6
Now you have 1/2x+8=1/4x
In order to subtract any fraction from another fraction you have to make sure they have a common denominator. In this case, the least common denominator between 2 and 4 would be 8, so now you have to multiply each denominator by the number that will make them equal 8.
1/2*4/4=4/8
1/4*2/2= 2/8
Now that you have a common factor, may the subtracting begin!
1/2x+8=1/4x becomes 4/8x+8 =2/8x
-2/8x -2/8x
2/8x=8 or 1/4x=8
Remember to divide by a fraction you need to multiply by its reciprocal so to solve for x you would need to multiply
4/1x*8/1=32
X=32
Plug in your solution to see if it’s correct
1/2(32) -6=1/4(32) +2
16-6=8+2
10=10 !
Single Variable Equations with Quadratics
If an equation has a quadratic in it, it means that there will be at least 2 solutions because quadratic functions have at least 2 x-intercepts.
Ex: X^2+6x+8=0
The first step is to factor the equation. We know that there will be two x’s inside the parenthesis. Now the only question is what two numbers added together will equal 6 and when multiplied together will equal 8. 2 and 4!
(x+2)(x+4)=0
So that means x=-2, x=-4
Since this function has two solutions you have to plug both of them in to see if they actually equal zero.
X^2+6x+8=0
(-2)^2+6(-2)+8=0
4+-12+8=0
-8+8=0 !
Now for the other solution
X^2+6x+8=0
(-4)^2+6(-4) +8=0
16-24+8
-8+8=0 !
*To check if they work you can also graph the parabola and make sure that both of these x-intercepts are equidistant from the vertex. Or can just have some fun with the quadratic equation*
Solving Single Variable Equations Witha Cube
Equations with a cube are similar to equations with a quadratic because you have to factor the equation first, but they are much more complicated because there are several different steps and many different methods of doing it.
Say you have X^3-6x^2+11x-6=0
We’ll solve this one using the factor theorem. The first thing you have to do is find the factors of -6 (the constant ) and x(the leading variable) and divide the factors of 6 by the factors of x using synthetic division and whichever number gives you a zero is a factor.
Factors of 6 are +1,-1, +2,-2, +3,-3, +6,-6
X=1, factors of one are +1 and -1
When you divide them you get: 1,-1, 2,-3, 3,-6 and 6 as the factors that you have to divide.
Then you perform synthetic division and your solutions turn out to be (x-1) (x-2) (x-3). So x=1, x=2 and x=3
Plug it into your equation to see if they actually turn out to be zero.
X^3-6x^2+11x-6=0
(1)^3-6(1)^2+11(1)-6
1-6+11-6= -5+5=0 !
Now for the second solution
(2)^3-6(2)^2+11(2)-6
8-24+22-6= -16+16=0!
Now for the last solution
(3)^3-6(3)^2+11(3)-6
27-54+33-6= -27+27=0!
*Another way to solve is to use long division to turn the equation into a quadratic and that will make it easier to factor. You can also graph the function and see where they cross the x- axis. Another way is to turn it into a form where there’s a factor and a quadratic, then just factor that quadratic to find the additional roots.*